Orthogonal System of Vectors
Table of Content |
In the orthogonal system of vectors we choose these vectors as three mutually perpendicular unit vectors denoted by 
, and 
directed along the
positive directions of X, Y and Z axes respectively.
Corresponding to any point P(x, y z) we can associate a vector w.r.t. a fixed orthogonal system and then this vector is the position vector (p.v.) of that point. i.e. p.v. of P = 
Distance of P from O = 
= 
x, y, z are called the components of the vector
makes angles a, b, g with the positive directions of X, Y and Z axes respectively, then cosa, cosb, cosg are called the direction cosines (d.c.’s) of 
cosb
; cosg = 
= 
.
Section Formula
Internal Division
Let A and B be two points with position vectors 
and 
respectively, and C be a point dividing AB internally in the ratio m : n. Then the position vector of C is given by 
.
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Proof: Let O be the origin. The
Þ n. Þ n(P.V. of Þ Þ |
|
, let 
) = m(P.V. of 
– P.V. of 
Þ 
or 
External Division
Let A and B be two points with position vectors and respectively and let C be a point dividing 
externally in the ratio m : n. Then the position vector of 
is given by
.
Note:
(i) If C is the mid–point of AB, then P.V. of C is 
.
(ii) We have,
. Hence 
is in the form of 
.
where, 
and 
.Thus, position vector of any point on can always be taken as 
where l + m = 1.
(iii) If circumcentre is origin and vertices of a triangle have position vectors 
, then the position vector of orthocentre will be 
.
Illustration 4: ABC is a triangle. A line is drawn parallel to BC to meet AB and AC in D and E respectively. Prove that the median through A bisects DE.
Solution: Take the vertex A of the triangle ABC as the origin. Let 
be the p.v. of B and C. The mid point of BC has the p.v. = 
The equation of the median is 
. Let D divide AB in the ratio 1:m
Þ p.v. of 
.Let E divide AC in the ratio 1:l
Þ p.v. E = 
Þl = m. p.v. of the mid-point of DE = 
which lies on the median. Hence the median bisects DE.
Isector of the Angle Between Two Vectors
Consider two non–zero, non–collinear vectors 
and 
. The bisector of the angle between the two vectors and is k
where k Î R+.
Illustration 5: If the vector 
bisects the angle between 
and 
, where is a unit vector then find .
Key concept: Bisector of the angle between the two vectors 
and 
is k
where k Î R+.
Solution: According to the given conditions l
= 
Þ 3
= 3l
= 
(3l + 1) – 
(2 + 9l) + 
(15l – 2)
Þ 
Þ 9 = (3l + 1)2 + (2 + 9l)2 + (15l – 2)2
Þ 315l2 – 18l = 0 Þl = 0, 
.
If l = 0, 
(not acceptable)
For l = , 
Scalar (or Dot) Product of Two Vectors
|
The scalar product of
Geometrical Interpretation: |
|
and 
, written as 
, is defined to be the 
cosq where q is the angle between the vectors 
is the product of length of one vector and length of the projection of the other vector in the direction of former.
. Projection of 
in direction of 
=.Projection of 
in direction of 
Properties:
Þ
(acosq)b = (projection of 
) b = (projection of 
) a
(Distributive law)
Û
are perpendicular to each other Þ
= 
If 
then
If 
are non-zero, the angle between them is given by 
= 
Illustration 6: Prove by vector method that (a1b1 + a2b2 + a3b3)2 £ (a12+ a22 + a32) (b12 + b22 + b32).
Solution: Let 
= a1
+ a2
+ a3 and 
= 
.
Now × = a1 b1 + a2b2 + a3b3 , also ×= || || cosq£ || ||.
Þ (×)2£ ||2 ||2 Þ (a1b1 + a2b2 + a3b3)2£
Illustration 7: If |
| = 3, |
| = 1, |
| = 4 and 
, find the value of 
Solution: We know, 
Þ 0 = 
(Given 
)
Þ0 = 
Þ
= –
= –13
Illustration 8: In a DABC, prove by vector method that
cos 2A + cos 2B + cos 2C ³ –3/2
Solution: As we know; 
…(i)
and 
… (ii)
Þ 3R2 + 2R2 (cos 2A + cos2B + cos 2C) ³ 0
Þ cos 2A + cos 2B + cos 2C ³ –3/2
Vector (Or Cross) Product of Two Vectors
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The vector product of two vectors |
|
, where q is the angle between the vectors 
If we curl the fingers of our right hand from to through the smaller angle (keeping the initial point of and same), the thumb points in the direction of 
. In this case, ,and (or 
), in that order are said to form a right handed system. It is evident that 
= absinq.
Properties:
-
Þ
-
(non-commutative) -
(Distributive) -
-
Û
are collinear (if none of 
is a zero vector) -
-
-
If
then 
-
=
- Any vector perpendicular to the plane of
is l () where l is a real number. Unit vector perpendicular to is ± 
|
Area is also treated as a vector with its direction in the proper sense. |
|
Illustration 9: and are unit vectors and || = 4. If angle between and is cos–1
and 
, then show that can be written as 
also find the value of l.
Key concept: If 
, then it can be written as 
=0, that means vector and (-2) are collinear.
Solution: Since vector and (-2
) are collinear, vector can be written as
Þ
Þ 
Þ16 + 4 – 4(4) (1) 
= l2(1)
Þ l2 = 16 Þl± 4.
Illustration 10: If , , are three vectors such that 
, 
, then show that 
= 1, 
Solution: Given 
…(i)
…(ii)
(1) – (2) Þ
Þ 
Now either 
is perpendicular to 
which is not possible
Or = 0 Þ
.
Also Þ
. 
.Equation the coefficients of , we get 
.
Thus and 
.
Scalar Triple Product
|
It is defined for three vectors i.e. V = magnitude of |
|
in that order as the scalar 
which can also be written simply as 
. It denotes the volume of the parallelopiped formed by taking a, b, c as the co-terminus edges.
The value of scalar triple product depends on the cyclic order of the vectors and is independent of the position of the dot and cross. These may be interchange at pleasure. However and anti-cyclic permutation of the vectors changes the value of triple product in sign but not a magnitude.
Properties:
-
If
are given as 
etc., then 
-
i.e. position of dot and cross can be interchanged without altering the product. Hence it is also represented by 
|
· · · |
|
-
in that order form a right handed system if 
Illustration 11: Show that 
.
Key concept: In such type of questions first we reduce each term as a product of two or three vector, by substituting a group of vectors by a single vector and then solve.
Solution: Let 
Now L.H.S. = 
= 
= 
= 
= 
= 0 = R.H.S.
Vector Triple Product
The vector product of two vectors, one of which is itself the vector product of two vectors, is a vector quantity called vector triple product.
It is defined for three vectors 
as the vector 
. This vector being perpendicular to 
, is coplanar with 
i.e. 
Take the scalar product of this equation with a. We get
0 = 
Þ
If we choose the coordinate axes in such a way that
, it is easy to show that l = 1. Hence
In general, 
( Vector triple product is not associative ) .
, if some or all of 
are zero vectors or 
are collinear.
Illustration 12: Let 
be three mutually perpendicular vectors of the same magnitude. If the vector 
satisfy the equation
then find
Solution: Here 
-
{
or l2
where 
l2 { 
+ 
then 
,
and 
.Þl2 {3
Þ3
Hence 
.
Collinear and Coplanar Vectors
1. Methods to Prove Collinearity
-
Two vectors
and 
are collinear if there exists kÎR such that 
. -
If
are collinear. -
Three points A(
), B(), C(
)are collinear if there exists kÎR such that 
that is 
. -
If
then A, B, C are collinear. -
A(
), B(), C() are collinear if there exists scalars l, m, n, (not all zero) such that 
where l + m + n = 0.
All the above methods are equivalent and any of them can be utilized to prove the collinearity. (of 2 vectors or 3 points)
Illustration 13: Let 
be three non–zero vectors such that any two of them are non–collinear. If 
is collinear with and 
is collinear with , then prove that 
.
Key concept: Two vectors 
and 
are collinear if there exists kÎR such that 
.
Solution: It is given that 
is collinear with
Þ
for some scalar l …(i)
Also 
is collinear with
Þ
for some scalar m …(ii)
from (i) and (ii)
Þ(1 + 2m) + (3 – ml) = 0
Þ1 + 2m = 0 and 3 – ml = 0 {
and are non–collinear vectors}
Þ m = – 1/2 and l = – 6
Substituting the values of l and m in (i) & (ii), we get
2. Methods to Prove Coplanarity
-
Three vectors
are coplanar if there exists l, mÎR such that 
i.e., one can be expressed as a linear combination of the other two. -
If
are coplanar (necessary and sufficient condition). -
Four points A
, B
, C
and D
lie in the same plane if there exist l, mÎR such that 
i.e. 
. -
If
= 0 then A, B, C, D are coplanar. -
A, B, C, D are coplanar if there exists scalars k, l, m, n (not all zero), such that
where k + l + m + n = o.
Again all the above methods are equivalent. Choose the best amongst them depending on convenience.
Illustration 14: Prove that if cos a¹ 1, cosb¹ 1 and cos g¹ 1, then the vectors 
can never be coplanar.
Key concept: If three vectorsare coplanar then 
.
Solution: Suppose that 
are coplanar.
Þ 
( R2® R2 – R1 and R3® R3 – R1 )
or 
or cosa (cosb – 1)(cosg – 1) – (1 – cosa)(cosg – 1) – (1 – cosa)(cosb – 1) = 0
dividing through out by (1 – cosa) (1 – cosb)(1 – cosg); we get
or –1 + 
Þ
, which is not possibleas 
Hence they can not be coplanar
Applications of Vectors
1. Vector Equation of a Straight Line
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Following are the two most useful forms of the equation of a line. (i) Line passing through a given point A |
|
where 
is the p.v. of any general point P on the line and l is any real number. The vector equation of a straight line passing through the origin and parallel to a vector is 
= n .
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(ii) Line passing through two given points A For each particular value of l, we get a particular point on the line. Each of the above equations can be written easily in Cartesian form also. For example, in case (i), writing
|
|
and B
: 
,
we get 
Þ x = a1 + lb1, y = a2 + lb2, z = a3 + lb3.
Illustration 15: Given vectors 
where O is the centre of circle circumscribed about DABC, then find vector 
.
Solution: Here, 
Þ
and 
|
Now Þ or Þ Þ |
|
[as radii of circle]
and 
…(i)
Now if we take 
then from (i),
…(ii)
and 
…(iii)
\ Solving (ii) and (iii);
and 
Þ 
2. Shortest Distacne Between Two Lines
Two lines in space can be parallel, intersecting or neither (called skew lines). Let 
be two lines.
(i) They intersect if 
.
(ii) They are parallel if 
are collinear. Parallel lines are of the form 
Perpendicular distance between them is constant and is equal to 
.
(iii) For skew lines, shortest distance between them (along common perpendicular) is given by 
.
3. Equation of a Plane in Vector Form
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Following are the four useful ways of specifying a plane. (i) A plane at a perpendicular distance d from the origin and normal to a given direction or
|
|
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(ii) A plane passing through the point A |
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has the equation 
(
.
(iii) Parameteric equation of the plane passing through Aand parallel to the plane of vectors 
is given by 
Þ.
(iv) Parameteric equation of the plane passing through A, B
C(A, B, C non-collinear) is given by 
Þ
.
In Cartesian form, the equation of the plane assumes the form Ax + By + Cz = D. The vector normal to this plane is 
and the perpendicular distance of the plane from the origin is 
.
Angle Between a Line and a Plane:
The angle between a line and a plane is the complement of the angle between the line and the normal to the plane.
Angle Between Two Planes:
It is equal to the angle between their normal unit vectors 
. i.e. cosq = 
4. Some Miscellaneous Result
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(i) Volume of the tetrahedron ABCD =
|
|
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(ii) Area of the quadrilateral with diagonals
|
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= 
.
5. Reciprocal System of Vectors
If 
are three non-coplanar vectors, then a system of vectors defined by 
is called the reciprocal system of vectors because 
.
Further 
The scalar product of any vector of one system with a vector of other system which does not correspond to it is zero i.e.
If 
is a reciprocal system to 
then is also reciprocal system to .
Illustration 16: Show that the points 
and 
are equi-distant from the plane r × (5i + 2j – 7k) + 9 = 0 and are on the opposite sides of it.
Solution: The given plane is 
= -9
Length of the perpendicular from 
to it is
Length of the perpendicular from 
= 
Thus the length of the two perpendiculars are equal in magnitude but opposite in sign. Hence they are located on opposite sides of the plane.
Solved Objective Problems
Q1. Any four non-zero vector will always be:
(A) Linearly dependent (B) Linearly independent
(C) Either ‘A’ or ‘B’ (D) None of these
Sol. Four or more than four non zero vectors are always linearly dependent.
Hence (A) is correct answer.
Q2. If 
and 
are reciprocal vectors, then:
(A) 
(B) 
(C) 
(D) None of these
Sol. If and are reciprocal, then
and 
Hence (C) is correct answer.
Q3. If 
and 
, then:
(A) 
(B) 
(C) 
(D) 
Sol. 
Also,
Thus 
Hence (C) is correct answer.
Q4. If three unit vectors 
satisfy 
, then angle between and is:
(A) 
(B) 
(C) 
(D) 
Sol. 
= 1
Hence (B) is correct answer.
Q5. Projection of 
on 
is equal to:
(A) 3 (B) -3
(C) 9 (D) -9
Sol. Projection of on is 
Thus required projection
= 
= 
Hence (B) is correct answer.
Q6. If and are two non-collinear unit vectors, then projection of 
on 
is equal to:
(A) 2 (B) -2
(C) 1 (D) None of these
Sol. 
Thus projection of on is zero.
Hence (D) is correct answer.
Q7. ABCD is a parallelogram with 
and 
. Area of this parallelogram is equal to:
(A) 
sq. units (B) 
sq. units
(C) 
sq. units (D) 
sq. units
Sol. Area vector of parallelogram
= 
= 
= 
= 
Area of the parallelogram
= 
sq. units
Hence (B) is correct answer.
Q8. If 
and 
always make an acute angle with each other for every value of 
, then:
(A) 
(B) 
(C) 
(D) 
Sol. 
= x(x + 1) + x – 1 + a
= 
+ 2x + a – 1
We must have
+ 2x + a – 1 > 0 
4 – 4(a – 1) < 0
a > 2
Hence (B) is correct answer.
Q9. Let be three non zero vectors such that Then 
, where 
is equal to:
(A) 1 (B) 2
(C) -1 (D) -2
Sol. Clearly 
and represents the sides of a triangle.
It’s area vector,
Thus, 
Hence (D) is correct answer.
Q10. If 
, 
where 
, then:
(A) 
(B) 
(C) 
(D)
Sol. 
Taking cross with in first equation, we get
and
Also, 
Hence (A) is correct answer.
Q11. Let be unit vectors such that 
, 
, 
, 
Then angle between and is :
(A) 
(B) 
(C) 
(D) 
Sol. 
Taking dot with on both sides, we get
If 
be the angle between and then
Hence (B) is correct answer.
Q12. Let be pair wise mutually perpendicular vectors, such that 
. Then length of 
is equal to:
(A) 2 (B) 4
(C) 3 (D) 6
Sol. 
= 
= 
= 1 + 4 + 4 + 0 + 0 + 0
= 9
= 3
Hence (C) is correct answer.
Q13. Let be three unit vectors such that 
Then 
is equal to:
(A) 
(B) 
(C) 
(D) None of these
Sol.
Thus 
are coplaner.
Hence 
Hence (D) is correct answer.
Q14. ABCD is a parallelogram 
and 
are the midpoints of side BC and CD respectively. If 
, then is equal to:
(A) 1/2 (B) 1
(C) 3/2 (D) 2
Sol. Let P.V. of A,B,D be 
and 
respectively.
Then P.V. of C = 
Also, P.V. of 
and, P.V. of 
Hence (C) is correct answer.
Q15. Two constant force 
and 
act on a particle. If particle is displaced from a point A with position vector 
to the point B with position vector 
Then work done in the process is equal to:
(A) 15 units (B) 10 units
(C) -15 units (D) -10 units
Sol. Total force, 
Displacement, 
Work done = 
= 2 – 12 – 5 = - 15 units.
Hence (C) is correct answer.
Q16. A, B, C and D are any four points in the space. If 
, where 
is the area of triangle ABC, then is equal to:
(A) 2 (B) 1/2
(C) 4 (D) 1/4
Sol. Let P.V. of A, B, C and D be and 
,
and 
= 
= 
= 
= 2(
)
= 4
= 4
Hence (C) is correct answer.
Q17. The position vector of the points A, B, C and D are 
, 
, 
and 
. It is known that these points are coplanar, then is equal to:
(A) 
(B) 
(C) 
(D) None of these
Sol. 
If vector 
and 
are coplanar, then
Hence (A) is correct answer.
Q18. Let be any three vectors. Then 
is always equal to:
(A) 
(B) 
(C) Zero (D) None of these
Sol. 
Hence (B) is correct answer.
Q19. Let be any three vectors. Then 
is also equal to:
(A) 
(B)
(C) Zero (D) None of these
Sol. 
Hence (A) is correct answer.
Q20. 
is always equal to:
(A) (B) 
(C) 
(D) 
Sol. 
Similarly, 
and 
Hence (C) is correct answer.
Q21. Value of 
is always equal to:
(A) 
(B) 
(C) 
(D) None of these
Sol.
Hence (A) is correct answer.
Q22. For any four vectors 
the expression 
is always equal to:
(A) 
(B) 
(C) 
(D) None of these
Sol. 
= 0
Hence (D) is correct answer.
Q23. For any four vectors and 
is always equal to:
(A) 
(B) 
(C) (D) 
Sol.
Hence (B) is correct answer.
Q24. In the parallelogram ABCD if the internal bisectors of the angle 
and 
intersect at the point P, then 
is equal to:
(A) (B)
(C) 
(D) 
Sol. Let P.V. of B, A and C be 
and respectively.
Now, 
and 
Hence (D) is correct answer.
Q25. If the vector 
bisects the angle between and 
, where is a unit vector, then:
(A) 
(B) 
(C) 
(D)
Sol. We must have
For 
(not acceptable)
For 
Hence (D) is correct answer.
Q26. Distance of 
from the plane 
is:
(A) 
(B) 
(C) 
(D) None of these
Sol. Let Q(
) be the foot of altitude drawn from P to the plane 
= 0,
Also 
Required distance
Hence (C) is correct answer.
Q27. Distance of from the line 
is:
(A) 
(B) 
(C) 
(D) None of these
Sol. (A) Let Q() be the foot of altitude drawn from P(
) to the line =
and 
Q28. Let and 
be unit vector that are mutually perpendicular, then for any arbitrary 
:
(A) 
(B) 
(C) 
(D) None of these
Sol. Let 
Also, 
and 
Hence (A) is correct answer.
Q29. The line 
will not meet the plane 
, provided:
(A) 
(B) 
(C) 
(D) 
Sol. We must have
and 
Hence (C) is correct answer.
Q30. The plane will contain the line , provided:
(A) (B)
(C) (D)
Sol. We must have
and 
Hence (C) is correct answer.
Q31. If the projection of point on the plane is the point 
, then:
(A) 
(B) 
(C) 
(D) 
Sol. We have
and 
Hence (B) is correct answer.
Q32. Let and be unit vectors that are perpendicular to each other, then 
will always be equal to:
(A) 1 (B) Zero
(C) -1 (D) None of these
Sol. 
= 
Hence (A) is correct answer.
Q33. If 
, then the vector 
is always equal to:
(A) (B)
(C) (D)
Sol. 
Hence (D) is correct answer.
Q34. For any two vectors and , the expression 
is always equal to:
(A) 
(B) 
(C) Zero (D) None of these
Sol. 
Similarly, 
and, 
Let 
Hence (B) is correct answer.
Q35. Let 
and 
be three non zero vectors such that 
, angle between and is and is perpendicular to and , then
where is equal to:
(A) 1/2 (B) 1/4
(C) 1 (D) 2
Sol. 
Hence (A) is correct answer.
Q36. Let be three vectors such that ,, 
Then:
(A) 
(B) 
(C) 
(D) 
Sol. 
Hence (A) is correct answer.
Q37. Let and be unit vectors such that 
, then the value of 
is equal to:
(A) 
(B) 
(C) 
(D) 
Sol.
Now, 
Hence (C) is correct answer.
Q38. Let be three unit vectors such that 
If the angle between and is , then 
, where 
is equal to:
(A) 
(B) 
(C) 
(D) None of these
Sol.
Hence (B) is correct answer.
Q39. Let P is any arbitrary point on the circumcircle of a given equilateral triangle of side length 
units then, 
is always equal to:
(A) 
(B) 
(C) 
(D) 
Sol. Let P.V. of P, A, B and C are 
and respectively and O() be the circumcentre of the equilateral triangle ABC.
Now, 
Similarly, 
and 
as 
Hence (A) is correct answer.
Q40. 
. A vector coplanar with and , whose projection on is of magnitude 
is:
(A) 
(B) 
(C) 
(D) 
Sol. Let the required vector be
Then, 
and 
Now, 
(2 – 2 – 1) + 
(2 – 1 – 2)
or 2
If 
, then
where 
If 
then
Hence (A) is correct answer.
Q41. If am + bm + cm, m =1, 2, 3, are pairwise perpendicular unit vectors, then 
is equal to
(A) 0 (B) 1 or –1
(C) 3 or -3 (D) 4 or –4
Sol. 
= 
=1 Þ
= ± 1 .
Hence (B) is correct answer.
Q42. If 
are three non-coplanar unit vectors, then
is equal to
(A) 
(B) 
(C) 
(D) 
Sol. 
= projection of in the direction of 
.
Hence the given vector is
Hence (D) is correct answer.
Q43. If sec2A
and
are coplanar, then cot2A + cot2B + cot2C is
(A) equal to 1 (B) equal to 2
(C) equal to 0 (D) not defined
Sol. The vectors are co-planar
Þ 
= 0
Þ cot2A + cot2 B + cot2 C + 1 = 0 which is not possible.
Hence (C) is correct answer.
Q44. If a, b, c are three non - coplanar vectors and p, q, r are vectors defined by the relations 
r = 
then the value of expression (a + b).p + (b + c).q + (c + a).r is equal to
(A) 0 (B) 1
(C) 2 (D) 3
Sol. 
Hence the given scalar expression = 1 + 1 + 1 = 3.
Hence (D) is correct answer.
Q45. The value of |a ´ |2 + |a ´ |2 + |a ´|2
is
(A) a2 (B) 2a2
(C) 3a2 (D) none of these
Sol. 
Þ |a|2 sin2a + |a|2 sin2b + |a|2 sin2g
= 3|a2| – |a2|(cos2a + cos2b + cos2g)
= 2|a2| = 2a2
Hence (B) is correct answer.
Q46. If 
are non-coplanar vectors then 
is equal to
(A) 3 (B) 0
(B) 1 (D) none of there
Sol. 
= 1 – 1 = 0.
Hence (B) is correct answer.
Q47. Consider DABC and DA1B1C1 in such a way that 
and M, N, M1 , N1 be the mid points of AB, BC, A1B1 and B1C1 respectively, then
(A) 
(B) 
(C) 
(D) 
Sol.
Þ
Þ 
Þ
Þ
Þ
Þ
Þ 2
Þ
Þ 2
Þ 
.
Hence (D) is correct answer.
Q48. Let 
, 
, where x1, x2, x3Î {-3, -2, -1, 0, 1, 2}. Number of possible vectors such that 
are mutually perpendicular, is
(A) 25 (B) 28
(C) 22 (D) none of these
Sol. 
Þ x1 + x2 + x3 = 0
Thus we have to obtain the number of integral solution of this equation.
Coefficient of x° | ( x-3 +x-2 +x-1 + x0 + x + x2)3
= x°
= x9 |(1 – x6)3 ( 1– x)-3 =11C9 – 3.5C3 = 25.
Hence (A) is correct answer.
Q49. Let a, b, c, be distinct and non-negative. If the vectors ai + aj + ck, i + k, and ci + cj + bk lie in a plane, then c is
(A) A.M. of a and b (B) G.M. of a and b
(C) H.M of a and b (D) equal to zero.
Sol. 
= 0
C2® C2 – C1
–1(ab – c2) = 0 Þ c2 = ab
Hence (B) is correct answer.
Q50. If 
, and 
,then 
is equal to
(A) 320 (B) 320
(C) - 320
(D) -320
Sol. 
and process gives on
= –320.
Hence (C) is correct answer.
Q51. If 
is the vector whose initial point divides the joining of 
and 
in the ratio k:1 and terminal point is origin. Also 
then the interval in which k lies
(A) (–¥, –6] È [–1/6, ¥) (B) (–¥, –6] È [1/6, ¥)
(C) (–¥, 6] È [–1/6, ¥) (D) (¥, 6] È [–1/6, ¥)
Sol. The point that divides 
and 
in the ratio of k : 1 is given
by 
Also 
Þ
Þ
On squaring both sides, we get 
Or 12k2 + 74k + 12 ³ 0Þ (6k + 1) (k + 6) ³ 0
Hence k Î (–¥, –6] È [–1/6, ¥).
Hence (A) is correct answer.
Q52. If 'a' is real constant and A, B, C are variable angles and, 
then the least value of 
is :
(A) 10 (B) 11
(C) 12 (D) 13
Sol: The given relation can be re–written as
Þ 
(as, a.b = |a| |b| cosq)
Þ 
Þ 
… (i)
also, 
(as, 
) … (ii)
from (i) and (ii), 
Hence least value of 
.
Hence (C) is correct answer.
Q53. The vector 
and 
are collinear for
(A) unique value of x , 0 < x < p/6
(B) unique value of x , p/6 < x < p/3
(C) no value of x
(D) infinity many value og x, 0 < x < p/2
Sol: Since and are collinear, for some l, we can write 
.
Þ 
Þ 
Þ cosx = x
Here we will get only one unique value of x which belongs to 
Hence (B) is correct answer.
Q54. The vectors 
have their initial points at (1, 1), the value of l so that the vectors terminate on one straight line is
(A) 0 (B) 3
(C) 6 (D) 9
Sol: Since initial point of 
is 
, their terminal points will be 
, 
and 
. Now given all the vectors terminate on one straight line. Hence
Þ l1 = 1 and l = 9
Hence (D) is correct answer.
Q55. Given that is a perpendicular to and p is a non-zero scalar, then a vector 
satisfying
is given by
(A) 
(B) 
(C) 
(D) none of these
Sol: We have 
. Taking dot by vector , we get
Þ 
Þ 
Þ 
Þ
.
Hence (A) is correct answer.
Q56. Let P is any arbitrary point on the circumcircle of a given equilateral triangle of side length '
' units then 
is always equal to:
(A) 
(B) 
(C) 
(D) 
Sol: Let P.V. of P, A, B and C are 
and respectively and O(
) be the circumcentre of the equilateral triangle ABC.
Þ 
Now 
Similarly, 
and 
Þ 
= 
Hence (A) is correct answer.
Q57. Let and are two non collinear vector such that 
. The angle of a triangle whose two sides are represented by the vector 
and 
are
(A) 
(B) 
(C) 
(D) none of these
Sol: Let
, clearly 
and 
are mutually perpendicular as is coplanar with and and is at right angle to the plane of and . And 
Þ
= 
= 
Also, 
= 
= 
Þ 
Thus angles are 
Hence (B) is correct answer.
Q58. E and F are the interior points on the sides BC and CD of a parallelogram ABCD. Let 
and 
. If the line EF meets the diagonal AC in G then 
where l is equal to
(A) 
(B) 
(C) 
(D) 
Sol: (D) Let P.V. of A.B. and D be 
. Then 
Þ 
and 
Equation of EF : 
Equation of AC : 
For point G we must have, 
Þ 
Þ
Hence (D) is correct answer.
Q59. If 
and the vector 
, 
and 
are non-coplanar, then the vectors 
and 
are:
(A) coplanar (B) none coplanar
(C) collinear (D) non collinear
Sol: Given 
[since X, Y, Z are non-coplanar]
Hence 
and 
are coplanar.
Hence (A) is correct answer.
Q60. If b and c are any two perpendicular unit vectors and a is any vector, then 
(A) b (B) a
(C) c (D) b + c
Sol: Consider three non-coplanar vectors b, c and 
. Any vector a can be written as 
……(i).
Taking dot product with in (i) we get
.Taking dot product with b in (i)
Taking dot product with c in (i), we get, a.c = y
Thus 
Hence (B) is correct answer.
Q61. If the lines 
and 
intersect (t and s are scalars) then.
(A) 
(B) 
(C) 
(D) none of these
Sol: For the point of intersection of the lines
Þ 
Hence (B) is correct answer.
Q62. If and then
(A) 
(B) 
(C) 
(D) none of these
Sol: 
Also 
, and 
Þ 
are mutually perpendicular vectors.
Þ 
and 
Þ 
Hence (C) is correct answer.
Q63. The position vector of a point P is 
where x, y, z Î N and 
. If 
= 10, then the number of possible positions of P is
(A) 30 (B) 72
(C) 66 (D) 9C2
Sol: Given 
= 10 Þ x + y + z = 10, x, y, z ³1
The number of possible positions of P
= coefficient of x10 in (x + x2 + x3 + … )3
= coefficient of x7 in (1 – x)-3
= 3 + 7 – 1C7 = 9C7 = 9C2 = 36
Hence (D) is correct answer.
Q64. If vectors ax
and x
make an acute angle with each other, for all x Î R, then a belongs to the interval
(A) 
(B) ( 0, 1)
(C) 
(D) 
Sol: Since vectors make an acute angle with each other so their dot product must be positive i.e. ax2 – 10 ax + 6 > 0 " x Î R
Þ- ax2 + 10ax – 6 < 0 " x Î R Þ –a < 0 and 100a2< 24a
Hence (C) is correct answer.
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